Wave Equation
1-D Wave Equation
\[
\begin{gather}
y(x, t)
\\\\
\frac{\partial^2 y}{\partial t^{2}}=c^{2}\frac{\partial^2 y}{\partial x^{2}}
\end{gather}
\]
Separate Variable Method
wave equation is on a closed interval \([0, L]\)
assume that
\[
\begin{gather}
y(x, t) = X(x)T(t)
\end{gather}
\]
then
\[
\begin{gather}
X_{n}(x) = \sin{(\frac{n\pi x}{p})}
\\\\
T_{n}(x) = a_{n}\cos{(\frac{n\pi c}{p}t)}+b_{n}\sin{(\frac{n\pi c}{p}t)}
\end{gather}
\]
finally we have
\[
\begin{gather}
y(x, t) = \sum_{n=1}^{\infty}{y_{n}(x,t)} = \sum_{n =1}^{\infty}{X_{n}(x)\ T_n(t)}
\end{gather}
\]
zero initial velocity
\[y_{t}(x, 0) = 0\]
let \(f(x)\) be an initial displacement function
\[
\begin{gather}
f(x) = y(x, 0)
\end{gather}
\]
then we can get
\[
\begin{gather}
T_{n}(t) = a_{n} \cos{(\frac{n\pi c }{p}t)}
\end{gather}
\]
and finally,
\[
\begin{gather}
y(x, t) = \sum_{n=1}^{\infty}{a_{n}\sin(\frac{n\pi x}{p})\cos{\frac{n\pi c}{p}t}}
\end{gather}
\]
- notice that p is a half of period (\([-p, p]\))
in which \(a_{n}\) is the sine Fourier series coefficient of \(f(x) = y(x, 0) = \sum{a_{n}\sin(\frac{n\pi x}{p})}\)
\[
\begin{gather}
a_{n} = \frac{2}{p}\int_{0}^{p}{f(x)\sin{(\frac{n \pi x}{p})}dx}
\end{gather}
\]
zero initial displacement
\[y(x, 0) = 0\]
let \(g(x)\) be the initial velocity function
\[
\begin{gather}
g(x) = y_{t}(x, 0)
\end{gather}
\]
then we can get
\[
\begin{gather}
T_{n}(t) = b_{n}\sin{(\frac{n\pi c}{p}t)}
\end{gather}
\]
in which \(\frac{n\pi c}{p}b_{n}\) is the Fourier sine series coefficient of \(g(x)\)
\[
\begin{gather}
\frac{n \pi c }{p}b_{n} = \frac{2}{p}\int_{0}^{p}{g(x)\sin{(\frac{n \pi x}{p})}dx}
\end{gather}
\]
finally,
\[
\begin{align}
y(x, t) &= \sum_{n=1}^{\infty}{y_{n}(x, t) }
\\\\
&=\sum_{n=1}^{\infty}{b_{n}\sin{(\frac{n\pi x}{p})}\sin{(\frac{n\pi c t}{p})}}
\end{align}
\]
nonzero initial velocity and displacement
in the situation, assume
\[y(x, t) = y_{1}(x, t) + y_{2}(x, t)\]
which \(y_{1}(x, t)\) assume zero initial velocity, and that \(y_{2}(x, t)\) assume zero initial displacement
and that is same as what we have discussed above.
with External Force
consider wave equation in this form
\[
\begin{gather}
\frac{\partial^2 y}{\partial t^{2}} = \frac{\partial^2 y}{\partial x^{2}}+Ax
\end{gather}
\]
\(y(x, t)\) can not be separated.
so we let
\[
\begin{gather}
y(x, t) = Y(x, t) + \phi(x)
\end{gather}
\]
assume that \(Y(x, t)\) can be separated, which means we can get the answer with the method above.
代入 \(y(x, t)\), we can get
\[
\begin{gather}
\frac{\partial^2 Y}{\partial t^{2}} = \frac{\partial^2 Y}{\partial t^{2}}+\frac{\partial^2 \phi}{\partial t^{2}}+Ax
\end{gather}
\]
therefore
\[
\begin{gather}
\phi''(x) + Ax =0
\end{gather}
\]
Infinite Medium
for the wave equations which are define in \((-\infty, \infty)\)
as usual, denote the initial displacement as \(f(x)\) and the initial velocity as \(g(x)\)
first we have to know that the eigenvalue only exists when \(\omega \geq 0\)
\[
\begin{gather}
X_{\omega}(x) = a_{\omega}\cos{(\omega x)}+ b_{\omega }\sin{(\omega x)}
\\\\
T_{\omega}(t) = A_{\omega}\cos{(\omega ct)} + B_{\omega}\sin{(\omega ct)}
\end{gather}
\]
then the most important thing is that
\[
\begin{gather}
y(x, t) = \int_{0}^{\infty}{y_{\omega}(x, t) \ d\omega} = \int_{0}^{\infty}{X_{\omega}(x)T_{\omega}(t)\ d\omega }
\end{gather}
\]
Semi-infinite Medium
for the wave equations which are define in \([0, \infty)\), thus
\[
\begin{gather}
y(0, t) = 0 \text{ for } t > 0
\end{gather}
\]
as usual, denote the initial displacement as \(f(x)\) and the initial velocity as \(g(x)\)
\[
\begin{gather}
X(x) = a\cos{(\omega x)}+ b\sin{(\omega x)}
\\\\
T(t) = c_{\omega}\cos{(\omega ct)} + d_{\omega}\sin{(\omega ct)}
\end{gather}
\]
%%
since \(X(0) = 0\), thus
\[
\begin{gather}
X(x) = b\sin{(\omega x)}
\end{gather}
\]
%%
Finite V.S. Infinite Medium
|
Finite |
Infinite |
|
Fourier sine series |
Fourier integral |
|
\(\frac{n\pi}{p}\) |
\(\omega\) |
| \(X(x)\) |
\(\sin{(\frac{n\pi x}{p})}\) |
\(a\cos{(\omega x)}+b\sin{(\omega x)}\) |
| \(T(t)\) when \(v(0)=0\) |
\(a_{n}\cos{(\frac{n\pi c}{p}t)}\) |
\(A_{\omega}\cos{(\omega ct)}\) |
| \(T(t)\) when \(d(0)=0\) |
\(b_{n}\sin{(\frac{n\pi c}{p}t)}\) |
\(B_{\omega}\sin{(\omega c t)}\) |
Characteristics and d'Alembert's Solution
one special solution for these condition
\[
\begin{gather}
u_{tt} = c^{2}u_{xx} \quad -\infty < x < \infty, t > 0
\\\\
u(x, 0) = f(x) \quad -\infty < x < \infty
\\\\
u_{t}(x, 0) = g(x) \quad -\infty < x < \infty
\end{gather}
\]
背
\[
\begin{gather}
u(x, t) = \frac{1}{2}\big[f(x-ct) + f(x+ct)\big] + \frac{1}{2c} \int_{x-ct}^{x+ct} g(k) \ dk
\end{gather}
\]
Forced Wave Motion
\[
\begin{gather}
\text{for }\quad -\infty < x < \infty,\quad t > 0
\\\\
u_{tt} = c^{2}u_{xx}+F(x, t)
\\\\
u(x, 0) = f(x)
\\\\
u_{t}(x, 0) = g(x)
\end{gather}
\]
\[
\begin{gather}
u(x, t) = \frac{1}{2} \big[f(x-ct)+f(x+ct)\big]+\frac{1}{2c}\int_{x-ct}^{x+ct}{g(x)\ dx} + \frac{1}{2c}\iint_{\Delta}F(x, t)\ dx\ dt
\end{gather}
\]
\(\Delta\) is characteristics triangle
\[
\begin{gather}
x-ct = k_{1}
\\\\
x+ct = k_{2}
\end{gather}
\]
\(k_{1}\), \(k_{2}\) can be any real constants.
2-D Wave Equation
\[
\begin{gather}
\frac{\partial z}{\partial t^{2}}=c^{2} \left(\frac{\partial^2 z}{\partial x^{2}}+\frac{\partial^2 z}{\partial y^{2}}\right)
\end{gather}
\]